Find $\tan\left(\dfrac{17\pi}{12}\right)$ exactly using an angle addition or subtraction formula. Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{\sqrt{3}+1}{\sqrt{3}-1}$ (Choice B) B $\dfrac{\sqrt{3}+3}{\sqrt{3}-3}$ (Choice C) C $\dfrac{-\sqrt{2}+\sqrt{6}}{4}$ (Choice D) D $\dfrac{-1-\sqrt{3}}{2}$
Explanation: The strategy First, we should rewrite the given angle $\dfrac{17\pi}{12}$ as the sum or difference of two special angles. Then, we can use the tangent addition or subtraction identities in order to evaluate $\tan\left(\dfrac{17\pi}{12}\right)$. [How do we find the trigonometric value of a sum or difference?] Rewriting $\dfrac{17\pi}{12}$ We can rewrite $\dfrac{17\pi}{12}$ as follows. $\begin{aligned}\dfrac{17\pi}{12}&=\dfrac{8\pi}{12}+\dfrac{9\pi}{12}\\\\\\ &=\dfrac{2\pi}{3}+\dfrac{3\pi}{4}\end{aligned}$ In other words, $\dfrac{17\pi}{12}$ is the sum of the special angles $\dfrac{2\pi}{3}$ and $\dfrac{3\pi}{4}$. Evaluating $\tan\left(\dfrac{17\pi}{12}\right)$ Using the tangent addition identity, we get the following. $\begin{aligned} \tan\left(\dfrac{17\pi}{12}\right)&= \tan\left(\dfrac{2\pi}{3}+\dfrac{3\pi}{4}\right) \\\\\\ &= \dfrac{\tan\left(\dfrac{2\pi}{3}\right)+\tan\left(\dfrac{3\pi}{4}\right)}{1-\tan\left(\dfrac{2\pi}{3}\right)\tan\left(\dfrac{3\pi}{4}\right)}\\\\\\ &= \dfrac{\left(-\sqrt{3}\right)+\left(-1\right)}{1-\left(-\sqrt{3}\right)\left(-1\right)}\\\\\\ &=\dfrac{\sqrt{3}+1}{\sqrt{3}-1} \end{aligned}$ Summary $\tan\left(\dfrac{17\pi}{12}\right) = \dfrac{\sqrt{3}+1}{\sqrt{3}-1}$